Win Hill: Inverse Marx Generator ??
in [Design]
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From: John Fields on 11 Jul 2010 14:54 >Good grief, you *don't* understand this stuff. > >You, like AlwaysWrong, are certainly smart enough to learn the basics >of electrical circuit math, but for some emotional reason you have >chosen not to. I see that a lot in techs. They compensate by attacking >people who can do the arithmetic, calling them eggheads or >"inexperienced" or argue over definitions and third-order effects to >obscure the fact that there *are* calculable answers. --- Typical Larkinese. I usually show my work, while you, on the other hand, are the one who always waits until close to the end of the thread to start "explaining" what everyone's already laid out, pretending that it was your answer in the first place and issuing gratuitous slurs in order to try to demean your detractors.
From: John Devereux on 11 Jul 2010 15:16 Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: > On Sun, 11 Jul 2010 18:26:35 +0100, John Devereux > <john(a)devereux.me.uk> wrote: > >>Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: >> >>> On Sun, 11 Jul 2010 09:05:34 +0100, John Devereux >>> <john(a)devereux.me.uk> wrote: >>> >>>>Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: >>>> >>>>> In the next few days, when I have time, I will issue a mathematical >>>>> proof that Larkin is totally wrong. Watch for it ;-) >>>>> >>>>> Why haven't Win Hill and Phil Hobbs come to Larkin's defense? >>>>> >>>>> Bwahahahaha! >>>> >>>>I'm no Phil Hobbs, but isn't all this argument because we are conflating >>>>two different usages of "charge"? >>>> >>>>The "charge" on a capacitor, as somone pointed out already, is really >>>>charge *separation* (dilectric polarization). The Q=CV refers to a >>>>*separation* of charge, not an absolute quantity. The "absolute" charge >>>>- the total number of electrons minus the number of protons - is >>>>normally low or zero. Unless your whole circuit picks up an >>>>electrostatic charge from somewhere else. It is this "absolute" charge >>>>which is conserved, the "Q=CV" "charge" of normal electronics is >>>>not. Take a solar cell charging a battery for one obvious example. As >>>>Larkin would say, where did the charge come from? Photons don't carry >>>>charge! >>> >>> The photons entered the game from "outside the box" as someone opined. >> >>But they don't carry charge. They do carry energy - so are you saying >>you can convert energy into "charge"? I agree - this is true for "Q=CV" >>"charge-as-used-in-electronics", but false for "number of electrons - >>number of protons" physicists charge. > > Yes, they carry energy and "knocked loose" electrons. Just to clarify what I assume is a typo there: they carry energy and "knock loose" electrons. That is, they do not carry electrons around with them! > But the total charge contained "inside the box" stays _constant_, > does it not? No charge was created, was it? Or destroyed? It really does depend what definition of charge you are using. A "differential" charge is created, Q=CV. The number of electrons and protons in the box stays constant. These are the only charge carrying particles, so the total "physicists" charge has to be constant. This is what is normally meant when we talk of laws of charge conservation. Yet not many *electronic engineers* would say that the "charge" in the battery/capacitor was zero. It *is* zero, strictly, physically, because the positive charge on the anode is cancelled by the negative charge on the cathode. But this is a useless definition for electronics - its pretty much *always* zero! The Q=CV "charge" that we talk about, in the electronics engineering context of capacitors, charge pumps and circuits, is not subject to "conservation". [...] -- John Devereux
From: The_Giant_Rat_of_Sumatra on 11 Jul 2010 15:26 On Sun, 11 Jul 2010 09:05:34 +0100, John Devereux <john(a)devereux.me.uk> wrote: >Jim Thompson <To-Email-Use-The-Envelope-Icon(a)On-My-Web-Site.com> writes: > >> In the next few days, when I have time, I will issue a mathematical >> proof that Larkin is totally wrong. Watch for it ;-) >> >> Why haven't Win Hill and Phil Hobbs come to Larkin's defense? >> >> Bwahahahaha! > >I'm no Phil Hobbs, but isn't all this argument because we are conflating >two different usages of "charge"? > >The "charge" on a capacitor, as somone pointed out already, is really >charge *separation* (dilectric polarization). The Q=CV refers to a >*separation* of charge, not an absolute quantity. The "absolute" charge >- the total number of electrons minus the number of protons - is >normally low or zero. Unless your whole circuit picks up an >electrostatic charge from somewhere else. It is this "absolute" charge >which is conserved, the "Q=CV" "charge" of normal electronics is >not. Take a solar cell charging a battery for one obvious example. As >Larkin would say, where did the charge come from? Photons don't carry >charge! I like this quote: Fudd's Law (Fudd's First Law of Opposition) "If you push something hard enough, it will fall over..." Essentially, it means "I think we're all Bozos on this bus."
From: The_Giant_Rat_of_Sumatra on 11 Jul 2010 15:31 On Sun, 11 Jul 2010 08:14:39 -0500, John Fields <jfields(a)austininstruments.com> wrote: >On Sun, 11 Jul 2010 13:47:20 +0100, John Devereux ><john(a)devereux.me.uk> wrote: > >>John Fields <jfields(a)austininstruments.com> writes: >> >>> On Sat, 10 Jul 2010 10:13:27 -0700, John Larkin >>> <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>> >>> >>>>OK, enlighten me. >>> >>> --- >>> OK. >>> --- >>> >>>>Slap a 1-ohm resistor across the 1F/1v cap and discharge it. You'll >>>>get 1 ampere-second out of it eventually. >>> >>> --- >>> Sorry, Charlie, but no. >>> >>> An ampere-second is the amount of charge transferred by a current of 1 >>> ampere in one second. >> >>That is, 1 coulomb. >> >>> >>> In your example the current will be one ampere when the resistor is >>> first connected, but will have decayed to about 368 mA after one >>> second has passed, so there's no way you'll get one ampere-second out >>> of it. >> >>What on earth are you talking about? This is pretty much the >>*definition* of capacitance. I.e., from Q = CV = "Ampere Seconds". >> >>No wonder John's having trouble convincing you of anything... > >--- >Not of anything, just of some things. > >About the ampere-seconds thing though: > >If you connect a 1VDC supply across a 1 ohm resistor for 1 second then >the amount of charge tranferred will be 1 coulomb. > >Then, since it got transferred in one second, the rate at which it was >tranferred was one coulomb per second, which is one ampere. > > >Now, replace the DC power supply with a capacitor charged to one volt, >connect it to the resistor, and then disconnect it after one second. > >Will one coulomb of charge have been transferred? Their brains decayed far faster than that cap after they left school.
From: The_Giant_Rat_of_Sumatra on 11 Jul 2010 15:33
On Sun, 11 Jul 2010 16:00:20 +0100, John Devereux <john(a)devereux.me.uk> wrote: > >Of course not, but Larkins original statement was "eventually", not >"after one second". You are the only one to add this condition, >obviously if you stop discharging the cap early you will not get all the >charge out! Herein lies the flaw with John L's claim, and your flaw if you are following his line of thinking. |